ATMO 551a Gravity Fall 2010 Gravity, Geopotential, Geoid and Satellite Orbits Much of this introductory material on gravity comes from Turcotte and Schubert (1982). The satellite orbit material comes from Elachi and van Zyl (2006). Gravitational acceleration The gravitational force between two point masses, M and m, pulling on one another is: GMm ˆ Fg = − 2 r r where G is the gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2 and r is the distance between the centers of the two masses in meters. The minus sign means the force of gravity pulls the masses toward€ one another. Since F = ma, the gravitational acceleration felt by mass, m, is simply a = GM/r2 (1) Consider Earth’s surface gravity. We assume at first that Earth is a perfect sphere so that it can be treated as a point mass. Given Earth’s mass =5.9736e24 kg and its average radius of 6.37101x106 m. The resulting surface gravity is 9.820660317 m/s2. This is quite close to the standard average surface gravity of 9.80665 m/s2. Note: GM is known far better than G or M. Issues related to Earth Spin… Earth is not a sphere The Earth gravitational acceleration we feel includes the effects of the Earth’s spin. There are two first order effects to be considered, the Earth is no longer a sphere and cannot be rtreated as a point mass and there is centrifugal force to be considered. Because of the spin of the Earth, the Earth has an equatorial bulge such that its equatorial radius is larger than its polar radius by about 20 km. This means a location at the pole is actually slightly closer to the Earth’s center than a location on the equator. Therefore we can anticipate that the surface gravity at the pole is slightly larger than the surface gravity at the equator (see Table below). The acceleration at the surface of Earth due to the force of gravity (which includes the effect of the greater radius at the equator) is GM 3GMa2J g = − + 2 3sin2 φ −1 r2 2r4 ( ) where a is Earth’s equatorial radius and φ is the latitude and J2 is the second spherical harmonic of Earth’s gravitational field (=1.08263e-3) which is a measure of Earth’s equatorial bulge. Notice in this €simple treatment there is no longitudinal dependence. In a more complex treatment of the gravity there is longitudinal dependence as well. Centrifugal force From the simple harmonic oscillator, we know the centrifugal force felt at the surface of the Earth due to the rotation of the Earth is 1 Kursinski 09/26/10 ATMO 551a Gravity Fall 2010 2 gω = ω s where ω is the angular velocity (=2 π /86164 seconds = 7.29e-5 radians/sec) and s is the perpendicular distance from the spin axis which is given as € s = r cos φ where r is the distance from the center of the Earth to the surface location being considered and φ is again the latitude. Therefore the centrifugal acceleration at Earth’s surface is 2 gω = ω rcosφ The magnitude of this centrifugal acceleration is largest at the equator where it is equal to 0.0339 m/s2. This is about 0.35% of the acceleration due to the force of gravity. The gravitational acceleration we feel at the surface€ defines the apparent local radial direction. So the component that is relevant to gravity is the radial component which is 2 2 gr '= gω cosφ = ω rcos φ Note that this is positive because it is an outward or upward acceleration so it decreases the acceleration due to the force of gravity. The full equation of€ gravity that includes the first order effect of spin is GM 3GMa2J g = − + 2 3sin2 φ −1 + ω 2rcos2 φ r2 2r4 ( ) I have compared this equation to a far more sophisticated equation of gravity derived from many years of satellite data and found it to be good to several parts in 104. The table€ below shows the contributions of the three terms. equatorial polar Units radius 6378.139 6356.7523 km latitude 0 1.570796 radians GM/r^2 -9.798280 -9.864322 m/s2 2 4 2 3GMa J2/2r -0.015912 0.032038 m/s centrifugal 0.033916 0 m/s2 Total gravity -9.780277 -9.832284 m/s2 So the equatorial gravity is indeed less than the polar gravity but not as much as the GM/r2 term would imply by itself. Sidereal day versus solar day (what’s my frame of reference) In order to calculate the spin rate of the Earth for the centrifugal term, we need to know the length of a day. What is the length of a day? This may seem like an obvious question but actually it is not and the answer depends on the frame of reference you choose. In what we normally call a year, the Earth rotates around the sun once per year. This number of days is actually one rotation less in comparison to the number of rotations relative to absolute space. A “sidereal” day is slightly shorter than a “solar day” because the Earth does not have to rotate quite as far to make a full revolution relative to absolute space as it does for the 2 Kursinski 09/26/10 ATMO 551a Gravity Fall 2010 sun to come directly overhead again. So there are 366.25 sidereal days per year compared to 365.25 solar days per year. So the length of a sidereal day is 365.25/366.25 * 86,400 sec = 86,164 sec. (Note there is a slight mistake in Elachi and Van Zyl on page 528). Points to a location in absolute space Points to the sun Change in gravity with altitude We can examine the radial dependence of gravity by performing a Taylor expansion of the gravitational acceleration around the value at the surface. GM 3GMa2J g = − + 2 3sin2 φ −1 + ω 2rcos2 φ r2 2r4 ( ) The vertical gradient of g is 2 ∂g GM 6GMa J2 2 2 2 € = +2 3 − 5 3sin φ −1 + ω cos φ ∂r r r ( ) The dominant term is the first term. Consider the fractional change in g with height. € 3 Kursinski 09/26/10 ATMO 551a Gravity Fall 2010 GM 6GMa2J −2 + 2 3sin2 φ −1 −ω 2 cos2 φ ∂g ∂ lng 3 5 ( ) = = r r g∂r ∂r GM r2 2 6a2J ω 2 cos2 φ = − + 2 3sin2 φ −1 − r r3 ( ) GM r2 r is approximately 6,371,000 m. Point mass term J2 term Centrifugal term Units -1 dlng/dr€ -3e-7 ~3e-9 -5e-10 m So at 10 km =104 m altitude the fractional change in g relative to the surface is 0.3%. For many applications this can be ignored. For some high precision applications, it cannot. Geopotential and the geoid The potential energy of an object in Earth’s gravity field can be determined by integrating the work done by the gravitational force in taking the object from an infinite distance to a finite distance r from Earth r' dW ΔW ∫ r' r' ⎡ GM 3GMa2J ⎤ ∞ gdr 2 3sin2 1 dr = = ∫ = ∫ ⎢− 2 + 4 ( φ − )⎥ m m ∞ ∞ ⎣ r 2r ⎦ ΔW GM GMa2J = − + 2 3sin2 φ −1 ≡ V (r) m r 2r3 ( ) The centrifugal€ term has not been included in the integral because it assumes a rigid rotation with the spin of the Earth which, at a distance of infinity, produces an infinite and unphysical rotational velocity.€ V is known as the gravitational potential which is the gravitational potential energy of a mass divided by its mass. A gravity potential, U, that accounts for both gravitation and rotation we can take the integral of the gravity equation: GM GMa2J 1 U(r) = − + 2 3sin2 φ −1 − ω 2r2 cos2 φ r 2r3 ( ) 2 The “equipotential” surface and why we care As you€ move along a surface on which the gravitational potential energy changes as you move along it, you will feel a force either pushing against you or accelerating you. Unlike a solid surface whose strength can oppose this force (so long as the force is not stronger than the solid’s strength), a fluid surface will respond to this force by moving and adjusting until its surface shape changes and there is no remaining horizontal force. Therefore (in the absence of other forces) the ocean surface is a surface along which the gravitational potential energy is constant. Such a surface is a called an equipotential surface. The reference equipotential surface that defines sea level is called the geoid. 4 Kursinski 09/26/10 ATMO 551a Gravity Fall 2010 The equatorial sea level geopotential (where φ = 0) is GM GMJ2 1 2 2 U(r = a) = − − − ω a ≡ U0 a 2a 2 The polar sea level geopotential (where φ=π/2) is GM GMa2J U r r 2 U € ( = pol ) = − + 3 ≡ 0 rpol rpol The flattening or ellipticity of this geoid is defined by a − r r f = pol =1− pol € a a rpol = a(1− f ) We assume that the Earth’s sea level surface is a geopotential and set the equatorial and polar geopotentials equal to get € € 2 GM GMJ2 1 2 2 GM GMa J2 − − − ω a = − + 3 a 2a 2 rpol rpol J 1 ω 2a3 a a3J a ⎛ a2J ⎞ 1 2 2 1 2 + + = − 3 = ⎜ − 2 ⎟ 2 2 GM rpol rpol rpol ⎝ rpol ⎠ € Subbing rpol = a(1− f ) and only using first order terms in f and J2 which are both very small yields € J 1 ω 2a3 a ⎛ a2J ⎞ 1 ⎛ J ⎞ 1+ 2 + = ⎜1 − 2 ⎟ = ⎜1 − 2 ⎟ € ⎜ 2 2 ⎟ ⎜ 2 ⎟ 2 2 GM a(1− f )⎝ a (1− f ) ⎠ (1− f )⎝ (1− f ) ⎠ J 1 ω 2a3 ⎛ J ⎞ 1 2 1 f ⎜1 2 ⎟ 1 f J + + = ( + )⎜ − 2 ⎟ = + − 2 2 2 GM ⎝ (1− f ) ⎠ € 3J 1 ω 2a3 f = 2 + 2 2 GM € -3 3 2 -3 Plugging in values of J2 = 1.08270x10 and a ω /GM = 3.46775x10 yields a value of f of 3.3579x10-3.